Question

Using the principal values of the inverse trigonometric functions the sum of the maximum and the minimum values of \(16((sec^{-1}x)^{2}+(cosec^{-1}x)^{2})\) is:

• A. \(24\pi^{2}\)
• B. \(18\pi^{2}\)
• C. \(31\pi^{2}\)
• D. \(22\pi^{2}\)

Hint:

Remember that the principal values of \(sec^{-1}x\) and \(cosec^{-1}x\) lie in specific intervals. Also, the sum of \(sec^{-1}x\) and \(cosec^{-1}x\) is \(\frac{\pi}{2}\).

Answer 1

The Correct Option is D

Step 1: Simplify the expression. Let \(sec^{-1}x = a\), where \(a \in [0, \pi] - \{\frac{\pi}{2}\}\). Then, \(cosec^{-1}x = \frac{\pi}{2} - a\). Substitute these into the given expression: \[ 16(sec^{-1}x)^{2}+(cosec^{-1}x)^{2} = 16[a^{2}+(\frac{\pi}{2}-a)^{2}] \] \[ = 16[a^{2} + \frac{\pi^{2}}{4} - \pi a + a^{2}] = 16[2a^{2} - \pi a + \frac{\pi^{2}}{4}] \] 
Step 2: Find the minimum value. To find the minimum value, take the derivative with respect to \(a\) and set it to zero: \[ \frac{d}{da}(2a^{2} - \pi a + \frac{\pi^{2}}{4}) = 4a - \pi = 0 \] \[ a = \frac{\pi}{4} \] Substitute \(a = \frac{\pi}{4}\) to find the minimum value: \[ min = 16[2(\frac{\pi}{4})^{2} - \pi (\frac{\pi}{4}) + \frac{\pi^{2}}{4}] = 16[\frac{2\pi^{2}}{16} - \frac{\pi^{2}}{4} + \frac{\pi^{2}}{4}] \] \[ min = 16[\frac{\pi^{2}}{8}] = 2\pi^{2} \] 
Step 3: Find the maximum value. The maximum value occurs at the endpoints of the interval for \(a\), which are \(a = 0\) and \(a = \pi\). At \(a = \pi\): \[ 16[2\pi^{2} - \pi (\pi) + \frac{\pi^{2}}{4}] = 16[2\pi^{2} - \pi^{2} + \frac{\pi^{2}}{4}] = 16[\frac{5\pi^{2}}{4}] = 20\pi^{2} \] At \(a = 0\): \[ 16[2(0)^{2} - \pi (0) + \frac{\pi^{2}}{4}] = 16[\frac{\pi^{2}}{4}] = 4\pi^{2} \] The maximum value is \(20\pi^{2}\). 
Step 4: Find the sum of the maximum and minimum values. \[ Sum = 2\pi^{2} + 20\pi^{2} = 22\pi^{2} \]